Question

The value of ${\int }_{-2}^0(x^3+{3x}^2+3x+3+(x+1\left)\cos \right(x+1\left)\right]dx$ is

• A. $0$
• B. $3$
• C. $4$
• D. $1$

Answer 1

The correct option is C $4$

$\begin{array}{c}{\int }_{-2}^0\left(x^3+3x^2+3x+3+x\cos cx+1\right)+\cos \left(x+1\right)]dx\\ \Rightarrow {\int }_{-2}^0{x}^{3}dx+3{\int }_{-2}^0{x}^{2}dx+3{\int }_{-2}^{0}xdx+3{\int }_{-2}^0\cos \left(x+1\right)dx+\left[x{\int }_{-2}^0\cos \left(x+1\right)dx\right]-\int \left[\frac{dx}{dx}.\int \cos \left(x+1\right)dx\right]dx\\ \Rightarrow \frac{x^4}{4}{\int }_{-2}^0+x^3{\int }_{-2}^0+\frac{3}{2}{\left[x^2\right]}_{-2}^0+{}^3{\left[x\right]}_{-2}^0+{\left[\sin \left(x+1\right)\right]}_{-2}^0+{\left[x\sin \left(x+1\right)+\cos \left(x+1\right)\right]}_{-2}^0\\ \Rightarrow \frac{-16}{4}+\frac{8}{2}-\frac{3}{2}\times 4+6+2\sin 1+\left[\cos 1-\left(2\sin 1+\cos 1\right)\right]\\ \Rightarrow 4+0+2\sin 1+\left[-2\sin 1\right]\\ =4\end{array}$